Timeline for Fundamental group of punctured simply connected subset of $\mathbb{R}^2$
Current License: CC BY-SA 4.0
18 events
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| Dec 17, 2019 at 5:56 | history | edited | Jeremy Brazas | CC BY-SA 4.0 |
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| Dec 17, 2019 at 3:51 | comment | added | George Lowther | ok, I added my proof, but it worked out a bit trickier than expected. | |
| Dec 17, 2019 at 3:10 | comment | added | Jeremy Brazas | Yes, something here will need to be fixed for the case when $S$ includes $\overline{B}(x,r)$ with other things attached. If $S$ is just a closed ball, the conclusion is pretty obvious. | |
| Dec 17, 2019 at 2:56 | comment | added | Thomas Browning | Why does there exist a point $y\in\overline{B}(x,r)\setminus S$? What if $S$ is a closed ball? | |
| Dec 17, 2019 at 2:40 | comment | added | Jeremy Brazas | @GeorgeLowther I feel like my argument is a fairly direct use of the Lemma but the difficulty I ran into is that you have no real control of what the image of a loop in $S\backslash\{x\}$ looks like or how it relates to $x$ and Lemma 13 requires a particular relationship between this image set and $x$. I found that it required a little finagling to make sure that $x$ actually ends up in the unbounded component of the inner region of a loop. | |
| Dec 17, 2019 at 2:27 | comment | added | George Lowther | I see you are using Lemma 13 already, so we are along the same lines, but I think it can be simplified. You need to show that a null-homotopic closed curve in $\mathbb R^2\setminus\{x\}$ and lying in $S$ is also null-homotopic in $S\setminus\{x\}$, which follows from the lemma. | |
| Dec 17, 2019 at 2:23 | comment | added | George Lowther | I haven't gone through your answer in detail, but I am aware of a reasonably direct proof using Lemma 13 from the quoted paper (as I used in this math.SE answer: math.stackexchange.com/a/3313163/1321) | |
| Dec 17, 2019 at 2:02 | comment | added | Jeremy Brazas | I apologize to everyone for editing this so much. This problem is far more interesting and delicate than I had anticipated. I'm cautiously hopeful that this new argument will either be a solution or lead to one. | |
| Dec 17, 2019 at 1:56 | history | edited | Jeremy Brazas | CC BY-SA 4.0 |
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| Dec 17, 2019 at 1:49 | history | edited | Jeremy Brazas | CC BY-SA 4.0 |
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| Dec 17, 2019 at 1:26 | history | edited | Jeremy Brazas | CC BY-SA 4.0 |
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| Dec 16, 2019 at 22:09 | history | edited | Jeremy Brazas | CC BY-SA 4.0 |
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| Dec 16, 2019 at 22:04 | history | edited | Jeremy Brazas | CC BY-SA 4.0 |
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| Dec 16, 2019 at 22:01 | comment | added | Neil Strickland | To elaborate on @YCor's first comment: if $G=\langle x,y|yx=xy^2\rangle$ then the pushout of $1\xleftarrow{}\langle x\rangle\to G$ is $1$, but $\langle x\rangle$ is an infinite cyclic group and a retract of $G$ as we see by sending $y$ to $1$. | |
| Dec 16, 2019 at 21:53 | comment | added | Jeremy Brazas | You are right, a more geometric argument is required. | |
| Dec 16, 2019 at 21:52 | comment | added | YCor | More precisely it is not true that given a path-connected space $X$ with a collared circle $C$ (i.e., with a closed subset homeomorphic to $C\times [0,1]$ with $C\times [0,1[$ being open in $X$, the "collared circle" being the copy of $C\times\{0\}$, assuming that gluing a disc along $C$ yields a simply connected space, does not imply that $X$ has cyclic fundamental group, and even if one assumes that $\pi_1(X)$ is torsion-free and that $\pi_1(C)$ is a retract subgroup of $\pi_1(X)$. | |
| Dec 16, 2019 at 21:22 | comment | added | YCor | My impression is that this argument only proves that the quotient of $\pi_1(U)$ by the normal closure of some cyclic (retract) subgroup is trivial. | |
| Dec 16, 2019 at 21:10 | history | answered | Jeremy Brazas | CC BY-SA 4.0 |